Problem: $\vec v = (6,-4)$ $\dfrac12\vec v= ($
In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $\dfrac12 \vec{v}$ : $\begin{aligned} {\dfrac12}\vec v = {\dfrac12} \cdot (6,-4) &= \left({\dfrac12} \cdot 6, {\dfrac12} \cdot (-4)\right) \\\\ &= (3,-2) \end{aligned}$ The answer is $ (3,-2) $.